(1 point) Rework problem 23 from section 2.4 of your text involving congressional committees. Assume that the committee consists of 11 Republicans and 6 Democrats. A subcommittee of 4 is randomly selected from all subcommittees of 4 which contain at least 1 Democrat.
What is the probability that the new subcommittee will contain at least 2 Democrats?

Step 1 ：Calculate the total number of ways to form a subcommittee of 4 from 11 Republicans and 6 Democrats. This is given by $(\genfrac{}{}{0ex}{}{17}{4})=2380$.

Step 2 ：Calculate the number of ways to form a subcommittee of 4 with at least 2 Democrats. This can be broken down into two cases:

Step 3 ：Case 1: The subcommittee has exactly 2 Democrats. There are $(\genfrac{}{}{0ex}{}{6}{2})$ ways to choose the 2 Democrats and $(\genfrac{}{}{0ex}{}{11}{2})$ ways to choose the 2 Republicans. So there are $(\genfrac{}{}{0ex}{}{6}{2})\times (\genfrac{}{}{0ex}{}{11}{2})=15\times 55=825$ ways for this case.

Step 4 ：Case 2: The subcommittee has more than 2 Democrats. This can be either 3 or 4 Democrats. For 3 Democrats, there are $(\genfrac{}{}{0ex}{}{6}{3})\times (\genfrac{}{}{0ex}{}{11}{1})=20\times 11=220$ ways. For 4 Democrats, there are $(\genfrac{}{}{0ex}{}{6}{4})=15$ ways (since all 4 members of the subcommittee are Democrats).

Step 5 ：Adding these up, there are $825+220+15=1060$ ways to form a subcommittee of 4 with at least 2 Democrats.

Step 6 ：The probability that the new subcommittee will contain at least 2 Democrats is therefore $\frac{1060}{2380}=\frac{53}{119}$.

Step 7 ：So, the final answer is $\boxed{{\displaystyle \frac{53}{119}}}$.